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Theorem elabg 2665
Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)
Hypothesis
Ref Expression
elabg.1 (x = A → (φψ))
Assertion
Ref Expression
elabg (A 𝑉 → (A {xφ} ↔ ψ))
Distinct variable groups:   ψ,x   x,A
Allowed substitution hints:   φ(x)   𝑉(x)

Proof of Theorem elabg
StepHypRef Expression
1 nfcv 2160 . 2 xA
2 nfv 1402 . 2 xψ
3 elabg.1 . 2 (x = A → (φψ))
41, 2, 3elabgf 2662 1 (A 𝑉 → (A {xφ} ↔ ψ))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98   = wceq 1228   wcel 1374  {cab 2008
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 617  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-10 1377  ax-11 1378  ax-i12 1379  ax-bnd 1380  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004
This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-nfc 2149  df-v 2537
This theorem is referenced by:  elab2g  2666  intmin3  3616  finds  4250  elxpi  4288  ovelrn  5572  peano5set  7309
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